The matrix $\left(I - J^\# J \right)$ is that of the projection onto $R(M^{-1}J^T)$, parallel to $\ker J$. It is not Hermitian, unless $R(J^T)$ (or equivalently $\ker J$) be stable under $M$. So, what do you mean by being non-negative definite ?
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Denis SerreJun 20 '12 at 10:14

Non-negative definite is equivalent to semi-positive definite, i.e., I would like to know if for an arbitrary vector $q$, the following relation holds: $q^T(I-J^{\#} J)q \ge 0$
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user24579Jun 20 '12 at 12:26

Modulo my comment above ($I-J^{\sharp}J$ is not Hermitian), here is what we can say:
First, it is correct that $MJ^\sharp J=J^T(JM^{-1}J^T)^{-1}J$ is Hermitian. So is $M(I-J^\sharp J)$.

Once you know that $M(I-J^\sharp J)$ is semi-definite positive, there follows that $I-J^\sharp J$ is diagonalizable with non-negative real eigenvalues. More generally, if $G,H$ are Hermitian with $H$ positive definite (here $G=M(I-J^\sharp J)$ and $H=M^{-1}$), then $HG$ is diagonalizable with real eigenvalues, and the signs of the eigenvalues of $HG$ agree with those of $G$.